Finding limit of a funciton with square roots.

[tmt1]

I have to find this:

$$\lim_{{x}\to{3}}\frac {\sqrt{6x - 14} - \sqrt { x+1}}{x-3}$$

So I do this:

$$\lim_{{x}\to{3}}\frac {\sqrt{6x - 14} - \sqrt { x+1}}{x-3} * \frac{\sqrt{6x + 14} + \sqrt{x+1}}{\sqrt{6x + 14} + \sqrt{x+1}}$$

The top part is easy since

$$(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) $$ is equal to $$a - b$$

So I have

$$\frac {5x- 14}{x\sqrt{6x+4} + x\sqrt{x+1} - 3\sqrt{6x+4} - 3\sqrt{x+1}} $$

From here I am lost.

 

[Evgeny.Makarov]

So I have

$$\frac {5x- 14}{x\sqrt{6x+4} + x\sqrt{x+1} - 3\sqrt{6x+4} - 3\sqrt{x+1}} $$

Then you divide the numerator and denominator by $x$ and find their limits separately.

Edit: Sorry, I missed the fact that the limit is when $x\to3$; I thought it is when $x\to\infty$.

 

[Dethrone]

Consider:

$$=\lim_{{x}\to{3}}\frac {\sqrt{6x - 14} - \sqrt { x+1}}{x-3}$$
$$=\lim_{{x}\to{3}}\frac {\sqrt{6x - 14}-2 - \sqrt { x+1}+2}{x-3}$$
$$=\lim_{{x}\to{3}}\frac {\sqrt{6x - 14}-2 - (\sqrt { x+1}-2)}{x-3}$$
$$=\lim_{{x}\to{3}}\frac{\sqrt{6x-14}-2}{x-3}-\frac{\sqrt{x+1}-2}{x-3}$$
$$=\d{}{x}\sqrt{6x-14}|_{x=3}-\d{}{x}\sqrt{x+1}|_{x=3}$$

The rest is easy :D

 

[soroban]

Hello, tmt!

Find: $\displaystyle \lim_{x\to3}\frac {\sqrt{6x - 14} - \sqrt { x+1}}{x-3}$

$\displaystyle \lim_{{x}\to{3}}\frac {\sqrt{6x - 14} - \sqrt { x+1}}{x-3} * \frac{\sqrt{6x - 14} + \sqrt{x+1}}{\sqrt{6x - 14} + \sqrt{x+1}}$

$\displaystyle \quad=\; \lim_{x\to3}\frac{(6x-14) - (x+1)}{(x-3)(\sqrt{6x-14} + \sqrt{x+1})} \;=\; \lim_{x\to3}\frac{5x-15}{(x-3)(\sqrt{6x-14} + \sqrt{x+1}}$

$\displaystyle \quad=\;\lim_{x\to3}\frac {5(x- 3)}{(x-3)(\sqrt{6x-14} + \sqrt{x+1}} \;=\; \lim_{x\to3}\frac{5}{\sqrt{6x-14} + \sqrt{x+1}}$

Can you finish it now?